Characteristic function and distribution of a random variable

Question

This is exercise 4.3 in Bjork, Arbitrage Theory in Continous Time. $$ X_t = \int^t_0 \sigma(s)dW_s $$ $\sigma$ is a deterministic function and $W_t$ is brownian motion.

I am asked to find the characteristic function of $X_t$ and thus showing that $X_t$ is normally distributed with mean zero and variance $\int^t_0 \sigma^2(s)ds$

I have found the characteristic function to be: $$ E[e^{iuX_t}]= \exp \left[-u^2/2 \int^t_0 \sigma^2(s)ds \right] $$ How Can I conclude that $X_t$ is normally distributed then?

Answer 1

The characteristic function (chf) defines the distribution function in a unique correspondence. For $X$ Gaussian with mean $0$ and variance $\sigma^2$ the chf $E[e^{i u X}]$ which is given by $$ e^{-\sigma^2 u^2/2}. $$ Thus if you identify the variance term then you are done. The characteristic function is the one of the Gaussian distribution. Thus the random variable $X_t$ is Gaussian.